In a survey of 400 likely voters, 215 responded that they would vote for the incumbent and 185 responded that they would vote for the challenger. Let p denote that fraction of all likely voters who...

In a survey of 400 likely voters, 215 responded that they would vote for the incumbent and 185 responded that they would vote for the challenger. Let p denote that fraction of all likely voters who preferred the incumbent at the time of the survey, and let pˆ be the fraction of survey respondents who preferred the incumbent.

a. At a 95% confidence level, can you conclude that the incumbent was ahead of the chal- lenger at the time of the survey? Explain.

In the survey being conducted of voters to gauge the outcome if an election was to be held, the accuracy of the result is reduced by the size of the sample. The sensitivity of the survey conducted in estimating the correct outcome can be arrived at as follows.

Take the desired sensitivity, here it is 95% or 0.95. Subtract this from 1, 1 - 0.95 = 0.05. Multiply the two values, 0.95*0.05 = 0.0475. Divide this by the sample size, here it is 500, 0.0475/400 = 1.1875*10^-4 . Take the square root of this value, `sqrt(1.1875*10^-4) ~~ 0.01` . For the 95% confidence interval the corresponding value in normal distribution tables is 1.96.

`sqrt(1.1875*10^-4)*1.96 ~~ 0.02135`

The value obtained above added and subtracted from 0.95 provides the 95% confidence interval of the survey. It is 0.97135 to 0.9286. 400 multiplied by these values gives 388 and 371.

185/371 = 0.4986 and 185/388 = 0.4768, in both the cases, the result is less than 0.5.

From the survey, it can be concluded with a 95% confidence interval that the incumbent was ahead of the challenger.